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# Ap calculus speeding problem

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### Transcript of "Ap calculus speeding problem"

1. 1. Warm up A. Find the average rate of change on [1, 2] of the function: f (x) = x3 - 2x B. Find the instantaneous rate of change when x = 2 of the same function.
2. 2. The Lame Joke of the day.. And now it’s time for.. Why did the scientist install a knocker on his door instead of a door bell? He wanted to win the No-bell prize
3. 3. The Lame Joke of the day.. And now it’s time for.. What is the formed by Ba + Na 2? Banana
4. 4. The Mean Value Theorem Lesson 4.2 I wonder how mean this theorem really is?
5. 5. This is Really Mean
6. 6. Think About It • Consider a trip of two hours that is 120 miles in distance …  You have averaged 60 miles per hour • What reading on your speedometer would you have expected to see at least once? 60
7. 7. Rolle’s Theorem • Given f(x) on closed interval [a, b]  Differentiable on open interval (a, b) • If f(a) = f(b) … then  There exists at least one number a < c < b such that f ’(c) = 0 f(a) = f(b) a bc
8. 8. Mean Value Theorem • We can “tilt” the picture of Rolle’s Theorem  Stipulating that f(a) ≠ f(b) • Then there exists a c such that a b c ( ) ( ) '( ) f b f a f c b a
9. 9. If f (x) is a differentiable function over [a,b], then at some point between a and b: f b f a f c b a Mean Value Theorem for Derivatives
10. 10. If f (x) is a differentiable function over [a,b], then at some point between a and b: f b f a f c b a Mean Value Theorem for Derivatives Differentiable implies that the function is also continuous.
11. 11. If f (x) is a differentiable function over [a,b], then at some point between a and b: f b f a f c b a Mean Value Theorem for Derivatives Differentiable implies that the function is also continuous. The Mean Value Theorem only applies over a closed interval.
12. 12. If f (x) is a differentiable function over [a,b], then at some point between a and b: f b f a f c b a Mean Value Theorem for Derivatives The Mean Value Theorem says that at some point in the closed interval, the actual slope (instantaneous rate of change) equals the average slope (average rate of change).
13. 13. y x 0 A B a b Slope of chord: f b f a b a Slope of tangent: f c y f x Tangent parallel to chord. c
14. 14. Example • Suppose you have a function f(x) = x^2 in the interval [1, 3]. ab afbf cf )()( )('cnumberaexiststherestatesMVTD 4 2 8 13 )1()3( :changeofrateaverage ff To find c, set derivative = to 4 xxf 2)(' 42x 2x
15. 15. 2][-2,intervalon the12)(functionheConsider t 3 xxxf ab afbf cf )()( )('cnumberastatesMVTD 8 22 )2()2( ff average 123)(' 2 xxf 8123)(' 2 ccf 3 42 c 155.1 3 2 c There can be more than 1 value of c that satisfy MVTD
16. 16. 2][-2,intervalon the 1 )(functionheConsider t x xf ab afbf cf )()( )('cnumberastatesMVTD There is no value of c that satisfies this equation. Why? f(x) is not continuous in the given interval!
17. 17. 17 Speeding problem Two police cars are located at fixed points 5 miles apart on a long straight road. • The speed limit is 55 mph • A car passes the first point at 53 mph • Four minutes later he passes the second car at 48 mph I think he was speeding We need to prove it Can the officers issue the man a ticket?
18. 18. Given f(x) = 5 – 4/x, find all c in the interval (1,4) such that the slope of the secant line = the slope of the tangent line. ? 14 )1()4( )(' ff cf 1 14 14 ?)(' xf 2 4 x 1 4 2 x 2 4 x 2x But in the interval of (1,4), only 2 works, so c = 2.
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